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Question: How do you simplify \(\tan ({\sin ^{ - 1}}(x))?\)...

How do you simplify tan⁡(sin⁡−1(x))?\tan ({\sin ^{ - 1}}(x))?

Explanation

Solution

Whenever such a type of question you have to solve, always assume ()() value to be x or θ, because it is always easy to solve. After that use Pythagorean Theorem.

Complete step by step solution:
As, mentioned in hint for solving such type of question, we will first assume sin⁡−1x=θ{\sin ^{ - 1}}x = \theta
So, sin⁡θ=x\sin \theta = x
Now we can rewrite the above equation as sin⁡θ=x1\sin \theta = \dfrac{x}{1}.
For 0<x<10 < x < 1, if we draw right triangle having Hypotenuse one and opposite x,
Then, sin⁡θ=OppositeHypotenuse\sin \theta = \dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}
Now applying Pythagorean Theorem for the right triangle having Hypotenuse one and opposite x, we get the other side of the triangle.
Which is 1−x2\sqrt {1 - {x^2}} .
Now, we already know that tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt {1 - {{\sin }^2}\theta }
Now substituting value of cos⁡θ\cos \theta in tan⁡θ\tan \theta equation we get,
tan⁡θ=sin⁡θ1−sin⁡2θ\tan \theta = \dfrac{{\sin \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}
Now, we will substitute value of θ in tan⁡θ\tan \theta equation,
So after substitute value of θ in tan⁡θ\tan \theta equation, we get,

tan⁡(sin⁡−1x)=x1−x2\tan ({\sin ^{ - 1}}x) = \dfrac{x}{{\sqrt {1 - {x^2}} }}

Note:
In such types of questions newer confuse or scare. If you just understand one question perfectly and after that practice one or two questions, you are able to solve any difficult question of such type. Just remember some basic trigonometry formulas and some basic rules. Always try to solve in a proper step by step method because if you have done any mistake there it is very difficult to find.